Advanced Higher Maths 2012-2013 : Discussion and Help Thread

Original post by helpmemaths

Does any one know how to do these questions:
Differentiate using the chain, product and quotient rules:

1) y=ln(3x+2)
2) y=(x+2)e^-x

For 1) use the chain rule.

For 2) use the product rule on (x+2) and e^-x

Reply 185

can anyone help me with this proof? can't get it it equal

k^2

..

\displaystyle \binom{k+2}{3} - \displaystyle \binom{k}{3} = k^2

Reply 186

Slumpy

16

Original post by Blue7195

can anyone help me with this proof? can't get it it equal

k^2

..

\displaystyle \binom{k+2}{3} - \displaystyle \binom{k}{3} = k^2

What have you done so far?
Write out the definitions, and let's see where you get to.

Reply 187

Original post by Slumpy

What have you done so far?
Write out the definitions, and let's see where you get to.

i just can't work out how to get the denominators to be the same.

\displaystyle \binom{k+2}{3} - \displaystyle \binom{k}{3}[br]= \dfrac{(k+2)!}{3!((k+2)-3)!} - \dfrac{3!}{3!(3-k)!}

Reply 188

Original post by Blue7195

i just can't work out how to get the denominators to be the same.

\displaystyle \binom{k+2}{3} - \displaystyle \binom{k}{3}[br]= \dfrac{(k+2)!}{3!((k+2)-3)!} - \dfrac{3!}{3!(3-k)!}

You should simplify the denominator of the left fraction first to 3!(k-1)!. You also have the right fraction being the definition of the upside-down binomial coefficient of the one you actually have.

To get the denominators the same, you should look for something you can multiply the smaller denominator by to get to the larger denominator, and then multiply top and bottom by that.

Reply 189

Slumpy

16

Original post by Blue7195

i just can't work out how to get the denominators to be the same.

\displaystyle \binom{k+2}{3} - \displaystyle \binom{k}{3}[br]= \dfrac{(k+2)!}{3!((k+2)-3)!} - \dfrac{3!}{3!(3-k)!}

Your second term there is wrong for one.

And looking at the first term, can you see any way to simplify (k+2)!/(k-1)! ?
(Obviously ((k+2)-3)!=(k-1)! )

Reply 190

Original post by ukdragon37

You should simplify the denominator of the left fraction first to 3!(k-1)!. You also have the right fraction being the definition of the upside-down binomial coefficient of the one you actually have.

To get the denominators the same, you should look for something you can multiply the smaller denominator by to get to the larger denominator, and then multiply top and bottom by that.

ahh thankyou, just have so many problems with these questions, can never work out what to multiply each fraction by

Reply 191

Original post by Slumpy

Your second term there is wrong for one.

And looking at the first term, can you see any way to simplify (k+2)!/(k-1)! ?
(Obviously ((k+2)-3)!=(k-1)! )

misplaced a 3! instead of a k! whoops :P

Reply 192

Original post by Blue7195

misplaced a 3! instead of a k! whoops :P

Also it should be (k-3) rather than (3-k)

Reply 193

Original post by ukdragon37

Also it should be (k-3) rather than (3-k)

thankyou!

Reply 194

Original post by Blue7195

thankyou!

So really, you would multiply the right fraction until the bottom became

3!(k-1)!

? the top would become

k!(k-2)(k-1)

? Then i could subtract the fractions? sorry for the questions, this part of the topic really confuses me

Reply 195

Original post by Blue7195

So really, you would multiply the right fraction until the bottom became

3!(k-1)!

? the top would become

k!(k-2)(k-1)

? Then i could subtract the fractions? sorry for the questions, this part of the topic really confuses me

Yep that's right :smile:

Reply 196

helpmemaths

Optimization problems

A sector of a circle with radius r cm has an area of 16cm²a) show that the perimeter P cm of the sector is given by

P(r)= 2(r+16/r)

b) Find the minimum value of P

Please help i have no idea how to do this question

Reply 197

Hype en Ecosse

20

Original post by helpmemaths

Optimization problems

A sector of a circle with radius r cm has an area of 16cm²a) show that the perimeter P cm of the sector is given by

P(r)= 2(r+16/r)

b) Find the minimum value of P

Please help i have no idea how to do this question

For part a) remember that the area of a sector

= \frac{\theta}{360} \pi r^2

and that the curvy bit on a sector =

\frac{\theta}{360} 2 \pi r

Can you see a substitution you can do to get rid of the part of the formula you don't have information on? :wink:

I'll give you a hint if get stuck

part b) Differentiate P(r) to find the values for which P'(r) = 0, you can use a nature table to find which is a minimum.

(edited 11 years ago)

Reply 198

Zepploydath

We were doing an optimisation question and it involved factorising 2ln(1/2x)-1=0

The final answer was x=0.303 but I'm not sure of the steps on how to get there.

Reply 199